Minimum Curve For the equation $y={{\left( x-p \right)}^{2}}+q$, • the graph opens upwards, • the

Question

Minimum Curve For the equation $y={{\left( x-p \right)}^{2}}+q$, • the graph opens upwards, • the coordinates of the minimum point of the graph are$\left( p,q \right)$ and • the graph is symmetrical about the line $x=p$. Maximum Curve For the equation $y=-{{\left( x-p \right)}^{2}}+q$, • the graph opens downwards, • the coordinates of the maximum point of the graph are $\left( p,q \right)$ and • the graph is symmetrical about the line $x=p$.

Accepted Answer

Both graphs have turning point $(p,q)$ and symmetry line $x=p$. The graph $y=(x-p)^2+q$ opens upwards with minimum $q$; the graph $y=-(x-p)^2+q$ opens downwards with maximum $q$.

Graphs of the form \(y=\pm {{\left( x-p \right)}^{2}}+q\)

Graphs of the form \(y=\pm {{\left( x-p \right)}^{2}}+q\)

Solution:

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